In a school meeting, 1/3 of people present are parents, 2/7 of people present are teachers, the remainder are…

Question

In a school meeting, 1/3 of people present are parents, 2/7 of people present are teachers, the remainder are pupils. Find the fraction of pupils present. If there are 64 pupils present, how many people are in the meeting? How many parents present?

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here are the solutions to the questions from the image.

Question 48: Father is 25 years older than his daughter. In 5 years' time, the father will be twice as old as the daughter.

a) How old is the daughter?

b) How old is the father?

c) How old will be the daughter?

d) How old will be the father?

Set up equations for current ages.
Let $D$ be the daughter's current age and $F$ be the father's current age.
From the first statement: $F = D + 25$ (Equation 1)
Set up equations for ages in 5 years.
In 5 years, daughter's age will be $D+5$ .
In 5 years, father's age will be $F+5$ .
From the second statement: $F+5 = 2(D+5)$ (Equation 2)
Solve the system of equations.
Substitute Equation 1 into Equation 2:
$(D+25) + 5 = 2(D+5)$ $D + 30 = 2D + 10$ $30 - 10 = 2D - D$ $D = 20$
Now find $F$ using Equation 1:
$F = 20 + 25 = 45$
Calculate ages in 5 years.
Daughter's age in 5 years = $D+5 = 20+5 = 25$ .
Father's age in 5 years = $F+5 = 45+5 = 50$ .

a) How old is the daughter?
The daughter is $\boxed{20 \text{ years old}}$ .

b) How old is the father?
The father is $\boxed{45 \text{ years old}}$ .

c) How old will be the daughter?
In 5 years, the daughter will be $\boxed{25 \text{ years old}}$ .

d) How old will be the father?
In 5 years, the father will be $\boxed{50 \text{ years old}}$ .

Question 49: Find the missing numbers in the sequences

a) 1, 4, 7, 10, ____, ____

b) 17, 19, 23, 29, ____, ____

c) 2, 3, 5, ____, ____

a) 1, 4, 7, 10, ____, ____
Identify the pattern.
This is an arithmetic sequence where each term is obtained by adding 3 to the previous term ( $4-1=3$ , $7-4=3$ , $10-7=3$ ).
Find the missing numbers.
The next number is $10+3=13$ .
The number after that is $13+3=16$ .
The missing numbers are $\boxed{13, 16}$ .

b) 17, 19, 23, 29, ____, ____
Identify the pattern.
The differences between consecutive terms are:
$19-17=2$
$23-19=4$
$29-23=6$
The differences are increasing by 2 each time (2, 4, 6).
Find the missing numbers.
The next difference should be $6+2=8$ . So, $29+8=37$ .
The next difference should be $8+2=10$ . So, $37+10=47$ .
The missing numbers are $\boxed{37, 47}$ .

c) 2, 3, 5, ____, ____
Identify the pattern.
This sequence consists of prime numbers in ascending order.
Find the missing numbers.
The prime numbers after 5 are 7 and 11.
The missing numbers are $\boxed{7, 11}$ .

Question 50: The cost price of a cow is $75000 \text{ Frw}$ and the selling price is $90000 \text{ Frw}$ . Calculate the percentage profit.
Calculate the profit.
$\text{Profit} = \text{Selling Price} - \text{Cost Price}$ $\text{Profit} = 90000 \text{ Frw} - 75000 \text{ Frw} = 15000 \text{ Frw}$
Calculate the percentage profit.
$\text{Percentage Profit} = \frac{\text{Profit}}{\text{Cost Price}} \times 100\%$ $\text{Percentage Profit} = \frac{15000 \text{ Frw}}{75000 \text{ Frw}} \times 100\%$ $\text{Percentage Profit} = \frac{1}{5} \times 100\% = 20\%$
The percentage profit is $\boxed{20\%}$ .

Question 51: The sum of three consecutive even numbers is 78. Find the numbers.
Represent the consecutive even numbers.
Let the first even number be $x$ .
The second consecutive even number is $x+2$ .
The third consecutive even number is $x+4$ .
Set up an equation and solve for $x$ .
$x + (x+2) + (x+4) = 78$ $3x + 6 = 78$ $3x = 78 - 6$ $3x = 72$ $x = \frac{72}{3} = 24$
Find the three numbers.
The first number is $x = 24$ .
The second number is $x+2 = 24+2 = 26$ .
The third number is $x+4 = 24+4 = 28$ .
The three consecutive even numbers are $\boxed{24, 26, 28}$ .

Question 52: A family eats $6 \text{ kg}$ of rice for 15 days. How many $\text{kg}$ of rice does the family eat in 6 days?
Calculate the daily consumption rate.
$\text{Daily consumption} = \frac{6 \text{ kg}}{15 \text{ days}} = \frac{2}{5} \text{ kg/day}$
Calculate the amount of rice eaten in 6 days.
$\text{Rice in 6 days} = \text{Daily consumption} \times 6 \text{ days}$ $\text{Rice in 6 days} = \frac{2}{5} \text{ kg/day} \times 6 \text{ days} = \frac{12}{5} \text{ kg}$ $\text{Rice in 6 days} = 2.4 \text{ kg}$
The family eats $\boxed{2.4 \text{ kg}}$ of rice in 6 days.

Question 53: Express $20 \text{ cm}$ as a fraction of $2 \text{ m}$ and simplify your answer.
Convert units to be consistent.
Since $1 \text{ m} = 100 \text{ cm}$ , then $2 \text{ m} = 2 \times 100 \text{ cm} = 200 \text{ cm}$ .
Form the fraction and simplify.
$\text{Fraction} = \frac{20 \text{ cm}}{200 \text{ cm}} = \frac{20}{200} = \frac{1}{10}$
The fraction is $\boxed{\frac{1}{10}}$ .

Question 54: On a map, $5 \text{ dm}$ represents $2 \text{ km}$ . Calculate the scale of the map.
Convert both measurements to the same unit, e.g., centimeters.
$1 \text{ dm} = 10 \text{ cm}$ , so $5 \text{ dm} = 5 \times 10 \text{ cm} = 50 \text{ cm}$ .
$1 \text{ km} = 1000 \text{ m} = 1000 \times 100 \text{ cm} = 100000 \text{ cm}$ .
So, $2 \text{ km} = 2 \times 100000 \text{ cm} = 200000 \text{ cm}$ .
Express the scale as a ratio and simplify to 1:N.
$\text{Scale} = \frac{\text{Map distance}}{\text{Actual distance}} = \frac{50 \text{ cm}}{200000 \text{ cm}}$ $\text{Scale} = \frac{50}{200000} = \frac{1}{4000}$
The scale of the map is $\boxed{1:4000}$ .

Question 55: James, Robert and Amos shared some money in the ratio of 4:3:9 respectively. If Amos got $3000 \text{ Frw}$ more than Robert.

a) How much money did Amos get?

b) How much money did James get?

c) How much money did Robert get?

d) How much money was shared?
Represent the shares using a common multiplier.
Let the common multiplier be $x$ .
James's share = $4x$
Robert's share = $3x$
Amos's share = $9x$
Use the given information to find $x$ .
Amos got $3000 \text{ Frw}$ more than Robert:
$9x - 3x = 3000 \text{ Frw}$ $6x = 3000 \text{ Frw}$ $x = \frac{3000}{6} = 500 \text{ Frw}$
Calculate each person's share.
a) How much money did Amos get?
Amos's share = $9x = 9 \times 500 \text{ Frw} = 4500 \text{ Frw}$ .
Amos got $\boxed{4500 \text{ Frw}}$ .

b) How much money did James get?
James's share = $4x = 4 \times 500 \text{ Frw} = 2000 \text{ Frw}$ .
James got $\boxed{2000 \text{ Frw}}$ .

c) How much money did Robert get?
Robert's share = $3x = 3 \times 500 \text{ Frw} = 1500 \text{ Frw}$ .
Robert got $\boxed{1500 \text{ Frw}}$ .

d) How much money was shared?
Total money shared = James's share + Robert's share + Amos's share
$\text{Total money} = 2000 \text{ Frw} + 1500 \text{ Frw} + 4500 \text{ Frw} = 8000 \text{ Frw}$
Alternatively, total parts = $4+3+9 = 16$ . Total money = $16x = 16 \times 500 \text{ Frw} = 8000 \text{ Frw}$ .
The total money shared was $\boxed{8000 \text{ Frw}}$ .

Question 56: Find the value of angles marked with letters X and Y.
Identify the angles in the triangle.
The interior angles of the triangle are $60^\circ$ , $Y$ , and $70^\circ$ .
Calculate angle Y using the sum of angles in a triangle.
The sum of angles in a triangle is $180^\circ$ .
$60^\circ + Y + 70^\circ = 180^\circ$ $130^\circ + Y = 180^\circ$ $Y = 180^\circ - 130^\circ = 50^\circ$
Calculate angle X.
Angle $X$ and the $60^\circ$ angle form a linear pair (angles on a straight line), so their sum is $180^\circ$ .
$X + 60^\circ = 180^\circ$ $X = 180^\circ - 60^\circ = 120^\circ$
The values of the angles are $\boxed{X = 120^\circ, Y = 50^\circ}$ .

Question 57: Work out and simplify completely

a) $\frac{1}{2} + \frac{1}{3} + \frac{1}{5}$

b) $(\frac{1}{4} + \frac{1}{2}) \div \frac{3}{4}$

c) $(3 - \frac{1}{3}) \times 13$

a) $\frac{1}{2} + \frac{1}{3} + \frac{1}{5}$
Find a common denominator for 2, 3, and 5, which is 30.
$\frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{1 \times 15}{2 \times 15} + \frac{1 \times 10}{3 \times 10} + \frac{1 \times 6}{5 \times 6}$ $= \frac{15}{30} + \frac{10}{30} + \frac{6}{30}$
Add the fractions.
$= \frac{15+10+6}{30} = \frac{31}{30}$
The simplified answer is $\boxed{\frac{31}{30}}$ .

b) $(\frac{1}{4} + \frac{1}{2}) \div \frac{3}{4}$
Add the fractions in the parenthesis.
$\frac{1}{4} + \frac{1}{2} = \frac{1}{4} + \frac{2}{4} = \frac{1+2}{4} = \frac{3}{4}$
Perform the division.
$\frac{3}{4} \div \frac{3}{4} = \frac{3}{4} \times \frac{4}{3} = 1$
The simplified answer is $\boxed{1}$ .

c) $(3 - \frac{1}{3}) \times 13$
Subtract the fractions in the parenthesis.
$3 - \frac{1}{3} = \frac{9}{3} - \frac{1}{3} = \frac{9-1}{3} = \frac{8}{3}$
Multiply the result by 13.
$\frac{8}{3} \times 13 = \frac{8 \times 13}{3} = \frac{104}{3}$
The simplified answer is $\boxed{\frac{104}{3}}$ .

Question 58: Write the following fractions in figures

a) Three tenths

b) Eleven and a quarter

c) Twelve thirty first

a) Three tenths
$\boxed{\frac{3}{10}}$
b) Eleven and a quarter
$\boxed{11\frac{1}{4}} \text{ or } \boxed{\frac{45}{4}}$
c) Twelve thirty first
$\boxed{\frac{12}{31}}$
Question 59: Simplify completely
$\frac{x^2 - 1}{x^2 + x}$
Factorize the numerator using the difference of squares formula ( $a^2 - b^2 = (a-b)(a+b)$ ).
$x^2 - 1 = (x-1)(x+1)$
Factorize the denominator by taking out the common factor $x$ .
$x^2 + x = x(x+1)$
Substitute the factored expressions back into the fraction and simplify.
$\frac{(x-1)(x+1)}{x(x+1)}$
Cancel out the common factor $(x+1)$ , assuming $x \neq -1$ .
$\frac{x-1}{x}$
The simplified expression is $\boxed{\frac{x-1}{x}}$ .

Question 60: If $\frac{2}{5}$ of the pupils in P6 class are girls. Find the number of boys in the class if there are 120 pupils.
Find the fraction of boys in the class.
The total fraction of pupils is 1.
$\text{Fraction of boys} = 1 - \text{Fraction of girls}$ $\text{Fraction of boys} = 1 - \frac{2}{5} = \frac{5}{5} - \frac{2}{5} = \frac{3}{5}$
Calculate the number of boys.
$\text{Number of boys} = \text{Fraction of boys} \times \text{Total pupils}$ $\text{Number of boys} = \frac{3}{5} \times 120$ $\text{Number of boys} = 3 \times \frac{120}{5} = 3 \times 24 = 72$
There are $\boxed{72}$ boys in the class.

Question 61: Senega, Kallas and Khadija contributed money to start a company. Senega paid $\frac{1}{2}$ of the cost and Kallas paid $\frac{1}{4}$ of the cost.

a) What fraction did Khadija contribute?

b) If Khadija contributed $200000 \text{ Frw}$ , what was the total cost?

a) What fraction did Khadija contribute?
Find the combined fraction contributed by Senega and Kallas.
$\text{Senega's fraction} + \text{Kallas's fraction} = \frac{1}{2} + \frac{1}{4}$ $= \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$
Find Khadija's fraction.
The total fraction of the cost is 1.
$\text{Khadija's fraction} = 1 - \frac{3}{4} = \frac{4}{4} - \frac{3}{4} = \frac{1}{4}$
Khadija contributed $\boxed{\frac{1}{4}}$ of the cost.

b) If Khadija contributed $200000 \text{ Frw}$ , what was the total cost?
Use Khadija's contribution and fraction to find the total cost.
Let the total cost be $C$ .
$\text{Khadija's fraction} \times C = \text{Khadija's contribution}$ $\frac{1}{4} \times C = 200000 \text{ Frw}$
Solve for $C$ .
$C = 200000 \text{ Frw} \times 4$ $C = 800000 \text{ Frw}$
The total cost was $\boxed{800000 \text{ Frw}}$ .

Question 62: Observe the following cylinder.
The image shows a cylinder with height $20 \text{ cm}$ and radius $7 \text{ cm}$ .

a) Calculate the total surface area of this cylinder.
Identify the given dimensions and the formula for the total surface area of a cylinder.
Height $h = 20 \text{ cm}$
Radius $r = 7 \text{ cm}$
The formula for the total surface area (TSA) of a cylinder is $TSA = 2\pi r^2 + 2\pi rh$ .
Use $\pi = \frac{22}{7}$ .
Substitute the values into the formula and calculate.
$TSA = 2 \times \frac{22}{7} \times (7 \text{ cm})^2 + 2 \times \frac{22}{7} \times 7 \text{ cm} \times 20 \text{ cm}$ $TSA = 2 \times \frac{22}{7} \times 49 \text{ cm}^2 + 2 \times 22 \times 20 \text{ cm}^2$ $TSA = 2 \times 22 \times 7 \text{ cm}^2 + 880 \text{ cm}^2$ $TSA = 308 \text{ cm}^2 + 880 \text{ cm}^2$ $TSA = 1188 \text{ cm}^2$
The total surface area of the cylinder is $\boxed{1188 \text{ cm}^2}$ .

In a school meeting, 1/3 of people present are parents, 2/7 of people present are teachers, the remainder are pupils. Find the fraction of pupils present. If there are 64 pupils present, how many people are in the meeting? How many parents present?

Step-by-step solution

Set up equations for current ages.

Set up equations for ages in 5 years.

Solve the system of equations.

Calculate ages in 5 years.

Identify the pattern.

Find the missing numbers.

Identify the pattern.

Find the missing numbers.

Identify the pattern.

Find the missing numbers.

Calculate the profit.

Calculate the percentage profit.

Represent the consecutive even numbers.

Set up an equation and solve for xxx.

Find the three numbers.

Calculate the daily consumption rate.

Calculate the amount of rice eaten in 6 days.

Convert units to be consistent.

Form the fraction and simplify.

Convert both measurements to the same unit, e.g., centimeters.

Express the scale as a ratio and simplify to 1:N.

Represent the shares using a common multiplier.

Use the given information to find xxx.

Calculate each person's share.

Identify the angles in the triangle.

Calculate angle Y using the sum of angles in a triangle.

Calculate angle X.

Find a common denominator for 2, 3, and 5, which is 30.

Add the fractions.

Add the fractions in the parenthesis.

Perform the division.

Subtract the fractions in the parenthesis.

Multiply the result by 13.

Factorize the numerator using the difference of squares formula (a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b)a2−b2=(a−b)(a+b)).

Factorize the denominator by taking out the common factor xxx.

Substitute the factored expressions back into the fraction and simplify.

Find the fraction of boys in the class.

Calculate the number of boys.

Find the combined fraction contributed by Senega and Kallas.

Find Khadija's fraction.

Use Khadija's contribution and fraction to find the total cost.

Solve for CCC.

Identify the given dimensions and the formula for the total surface area of a cylinder.

Substitute the values into the formula and calculate.

Still stuck on this one?

More Mathematics Questions

Set up an equation and solve for $x$ .

Use the given information to find $x$ .

Factorize the numerator using the difference of squares formula ( $a^2 - b^2 = (a-b)(a+b)$ ).

Factorize the denominator by taking out the common factor $x$ .

Solve for $C$ .